Chapter 14, Problem 48E

Interpretation Introduction

(a)

Interpretation:

Whether the statement “the molar volume of a gas at $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ is less than $\text{22}\text{.7}\text{L}/\text{mol}$” is true or false is to be stated.

Concept introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is represented as given below.

$\text{PV}=\text{nRT}$

Answer to Problem 48E

The statement “the molar volume of a gas at $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ is less than $\text{22}\text{.7}\text{L}/\text{mol}$” is false.

Explanation of Solution

It is given that pressure and temperature for the gas is $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ respectively.

The conversion factor of temperature unit $°\text{C}$ into $\text{K}$ is done below.

$\text{K}=°\text{C+273}$

Substitute the value of temperature in the above equation as shown below.

$\begin{array}{rll}\text{K}& =& °\text{C+273}\\ & =& 212+273\\ & =& 485\text{K}\end{array}$

Ideal gas equation is given by the formula as shown below.

$\text{PV}=\text{nRT}$…(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{T}$ is the temperature.

• $\text{n}$ is the number of moles.

• $\text{R}$ is the gas constant.

Rearrange the above equation in terms of molar volume as shown below.

$\begin{array}{rll}\text{PV}& =& \text{nRT}\\ \frac{\text{V}}{\text{n}}& =& \frac{\text{RT}}{\text{P}}\end{array}$…(2)

Substitute the values of pressure and temperature of gas in the equation (2) to calculate molar volume as shown below.

$\begin{array}{rll}\frac{\text{V}}{\text{n}}& =& \frac{\text{RT}}{\text{P}}\\ & =& \frac{\text{0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\times 485\text{K}}{\text{1}\text{.06}\text{atm}}\\ & =& 37.56\text{L}{\text{mol}}^{-1}\end{array}$

The molar volume of a gas at $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ is $37.56\text{L}{\text{mol}}^{-1}$ which is more than $\text{22}\text{.7}\text{L}/\text{mol}$.

Therefore, the statement “the molar volume of a gas at $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ is less than $\text{22}\text{.7}\text{L}/\text{mol}$” is false.

Conclusion

The statement “the molar volume of a gas at $\text{1}\text{.06}\text{atm}$ and $\text{212}\text{°C}$ is less than $\text{22}\text{.7}\text{L}/\text{mol}$” is false.

Interpretation Introduction

(b)

Interpretation:

Whether the statement “the mass of $\text{5}\text{.00}\text{L}$ of ${\text{NH}}_{\text{3}}$ is the same as the mass of $\text{5}\text{.00}\text{L}$ of $\text{CO}$ if both volumes are measured at the same temperature and pressure” is true or false is to be stated.

Concept introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is represented as given below.

$\text{PV}=\text{nRT}$

Answer to Problem 48E

The statement “the mass of $\text{5}\text{.00}\text{L}$ of ${\text{NH}}_{\text{3}}$ is the same as the mass of $\text{5}\text{.00}\text{L}$ of $\text{CO}$ if both volumes are measured at the same temperature and pressure” is false.

Explanation of Solution

Ideal gas equation is given by the formula as shown below.

$\text{PV}=\text{nRT}$…(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{T}$ is the temperature.

• $\text{n}$ is the number of moles.

• $\text{R}$ is the gas constant.

The number of moles is given by the formula as shown below.

$\text{n}=\frac{\text{Given}\text{mass}\left(\text{m}\right)}{\text{Molar}\text{mass}\left(\text{MM}\right)}$

Substitute the above equation in equation (1) as shown below.

$\text{PV}=\frac{\text{m}}{\text{MM}}\text{RT}$…(2)

Rearrange the above equation in terms of mass as shown below.

$\begin{array}{rll}\text{PV}& =& \frac{\text{m}}{\text{MM}}\text{RT}\\ \text{m}& =& \frac{\text{PV}\times \text{MM}}{\text{RT}}\end{array}$…(3)

From the above equation, it is clear that the mass of a gas is directly proportion to its molar mass. This means the mass of $\text{5}\text{.00}\text{L}$ of ${\text{NH}}_{\text{3}}$ is not the same as the mass of $\text{5}\text{.00}\text{L}$ of $\text{CO}$.

Therefore, the statement “the mass of $\text{5}\text{.00}\text{L}$ of ${\text{NH}}_{\text{3}}$ is the same as the mass of $\text{5}\text{.00}\text{L}$ of $\text{CO}$ if both volumes are measured at the same temperature and pressure” is false.

Conclusion

The statement “the mass of $\text{5}\text{.00}\text{L}$ of ${\text{NH}}_{\text{3}}$ is the same as the mass of $\text{5}\text{.00}\text{L}$ of $\text{CO}$ if both volumes are measured at the same temperature and pressure” is false.

Interpretation Introduction

(c)

Interpretation:

Whether the statement “at a given temperature and pressure, the densities of two gases are proportional to their molar masses” is true or false is to be stated.

Concept introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is represented as given below.

$\text{PV}=\text{nRT}$

Answer to Problem 48E

The statement “at a given temperature and pressure, the densities of two gases are proportional to their molar masses” is true.

Explanation of Solution

Ideal gas equation is given by the formula as shown below.

$\text{PV}=\text{nRT}$…(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{T}$ is the temperature.

• $\text{n}$ is the number of moles.

• $\text{R}$ is the gas constant.

The number of moles is given by the formula as shown below.

$\text{n}=\frac{\text{Given}\text{mass}\left(\text{m}\right)}{\text{Molar}\text{mass}\left(\text{MM}\right)}$

Substitute the above equation in equation (1) as shown below.

$\text{PV}=\frac{\text{m}}{\text{MM}}\text{RT}$…(2)

The density is given by the formula as shown below.

$\text{D}=\frac{\text{m}}{\text{V}}$

Rearrange equation (2) in terms of density using the above equation as shown below.

$\begin{array}{rll}\text{PV}& =& \frac{\text{m}}{\text{MM}}\text{RT}\\ \text{P}& =& \frac{\text{m}}{\text{V}\times \text{MM}}\text{RT}\\ \text{MM}& =& \frac{\text{D}\times \text{RT}}{\text{P}}\end{array}$…(3)

From the above equation, it is clear that the molar mass of a gas is directly proportion to its density.

Therefore, the statement “at a given temperature and pressure, the densities of two gases are proportional to their molar masses” is true.

Conclusion

The statement “at a given temperature and pressure, the densities of two gases are proportional to their molar masses” is true.

Interpretation Introduction

(d)

Interpretation:

Whether the statement “to change liters of a gas to moles, multiply by $\text{RT}/\text{P}$” is true or false is to be stated.

Concept introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is represented as given below.

$\text{PV}=\text{nRT}$

Answer to Problem 48E

The statement “to change liters of a gas to moles, multiply by $\text{RT}/\text{P}$” is false.

Explanation of Solution

Ideal gas equation is given by the formula as shown below.

$\text{PV}=\text{nRT}$…(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{T}$ is the temperature.

• $\text{n}$ is the number of moles.

• $\text{R}$ is the gas constant.

Rearrange equation (1) in terms of number of moles as shown below.

$\begin{array}{rll}\text{PV}& =& \text{nRT}\\ \text{n}& =& \frac{\text{PV}}{\text{RT}}\end{array}$…(2)

From the above equation, it is clear that to calculate number of moles from liters of a gas, the volume is multiplied by $\frac{\text{P}}{\text{RT}}$.

Therefore, the statement “to change liters of a gas to moles, multiply by $\text{RT}/\text{P}$” is false.

Conclusion

The statement “to change liters of a gas to moles, multiply by $\text{RT}/\text{P}$” is false.